Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $q = \dfrac{-4t - 36}{t^2 - 7t + 12} \div \dfrac{-9t - 81}{3t - 12} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-4t - 36}{t^2 - 7t + 12} \times \dfrac{3t - 12}{-9t - 81} $ First factor the quadratic. $q = \dfrac{-4t - 36}{(t - 4)(t - 3)} \times \dfrac{3t - 12}{-9t - 81} $ Then factor out any other terms. $q = \dfrac{-4(t + 9)}{(t - 4)(t - 3)} \times \dfrac{3(t - 4)}{-9(t + 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ -4(t + 9) \times 3(t - 4) } { (t - 4)(t - 3) \times -9(t + 9) } $ $q = \dfrac{ -12(t + 9)(t - 4)}{ -9(t - 4)(t - 3)(t + 9)} $ Notice that $(t + 9)$ and $(t - 4)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -12(t + 9)\cancel{(t - 4)}}{ -9\cancel{(t - 4)}(t - 3)(t + 9)} $ We are dividing by $t - 4$ , so $t - 4 \neq 0$ Therefore, $t \neq 4$ $q = \dfrac{ -12\cancel{(t + 9)}\cancel{(t - 4)}}{ -9\cancel{(t - 4)}(t - 3)\cancel{(t + 9)}} $ We are dividing by $t + 9$ , so $t + 9 \neq 0$ Therefore, $t \neq -9$ $q = \dfrac{-12}{-9(t - 3)} $ $q = \dfrac{4}{3(t - 3)} ; \space t \neq 4 ; \space t \neq -9 $